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The forward Euler method applied to the logistic differential equation $y^{\prime }=ry(1-ay)$ becomes the logistic map after $u=a(\lambda -1)y/\lambda$, $\lambda =1+r\Delta t$. The positive equilibrium point of the differential equation has perturbations decaying as $e^{-rt}$. Its discrete fixed-point multiplier is $1-r\Delta t=2-\lambda$: for $r\Delta t>2$, the discretization has an artificial oscillatory instability, leading initially to alternation on successive steps. This contrasts with the positive perturbation multiplier $e^{-r\Delta t}$ of the exact time-step map.

A density deficit in a flux tube is not by itself an instability calculation. For example, a straight magnetized tube in otherwise field-free gas at the same temperature has lateral balance $p_i+B_i^2/(2{\mu }_0)=p_e$. The perfect gas law then gives ${\rho }_i<{\rho }_e$, so the tube rises unless some other force balances its weight deficit. This configuration simply lacks magnetostatic equilibrium. A magnetic buoyancy instability, in contrast, starts from a force-balanced atmosphere and asks whether a small allowed displacement grows. A balanced horizontal magnetic layer can be stable or unstable; the presence of magnetic pressure alone does not decide this.

Assume a smooth equilibrium, with downward gravitational acceleration $g>0$, and write $\mathcal{R}$ for the gas constant per unit mass. The ambient total pressure and force balance areWe use ideal magnetic flux freezing on the displacement time scale and rapid lateral pressure equilibration. A bodily displacement perpendicular to a straight horizontal tube does not change its length or bend its field. If its area is $\mathcal{A}$, conservation of magnetic flux and mass per unit length gives $B\mathcal{A}=\mathrm{constant}$ and $\rho \mathcal{A}=\mathrm{constant}$. Thus the tube preservesEfficient heat exchange fixes its temperature to the ambient temperature at its new height, not to its original temperature. This is the thermal assumption that distinguishes the requested criterion from the adiabatic one.

Take a tube initially at $z_0$ and let $f_0=B_0(z_0)/\rho (z_0)$. After a small upward displacement $\xi$, let ${\rho }_i(z)$ be the density it would have in lateral balance at $z=z_0+\xi$. Its preserved $f_0$ and imposed local temperature giveAt the initial position ${\rho }_i=\rho$. Differentiate this relation at $z_0$:For the ambient atmosphere the same differentiation must also differentiate $f(z)=B_0(z)/\rho (z)$. It givesSubtracting isolates the actual displaced density contrast:The pressure-gradient force on the tube balances the weight of the displaced ambient gas, so its leading vertical buoyancy acceleration isThe coefficient multiplying $\xi$ is positive precisely when the tube becomes lighter on upward displacement. Thus the isothermal interchange criterion for magnetic buoyancy isWhere $B_0>0$, this is simply $d(B_0/\rho )/dz<0$, or $d\log B_0/dz<d\log \rho /dz$. The squared form is independent of the field's choice of direction. Equality is neutral at this order, and the reversed inequality is restoring for these displacements. A layer with an interval satisfying the strict inequality admits local growing interchange disturbances under the assumed pressure/thermal equilibration. The calculation does not address field-bending undular modes, finite heat-transfer times or magnetic diffusion; those can change the criterion. In particular, the localized interchange criterion for a magnetized atmosphere with an adiabatic gas response is not interchangeable with this fast-thermal-exchange result.

Let $g>0$ be the magnitude of the downward gravity and write ${\alpha }_T$ for the thermal expansion coefficient, to distinguish it from the dynamo coefficient of the previous question. The conductive state has $\mathbf{u}=0$, $\mathbf{B}=B_0\hat{\mathbf{z}}$ and $T_0(z)=T_r-\Delta Tz/d$. Its gravitational term is absorbed in a background hydrostatic pressure.

For the dimensional two-dimensional fields, choose the stream function and Cartesian magnetic flux function conventionsThese automatically give zero divergence. Define the Jacobian determinant bracket $J(f,h)=f_x{h}_z-f_z{h}_x$, so $\mathbf{u}\cdot \nabla h=J(\psi ,h)$. The full magnetic flux function, including the imposed field, must be retained in the magnetic force; using the perturbation alone would lose its linear restoring term.

The scalar vorticity is $(\nabla \times \mathbf{u}{)}_y=-{\nabla }^2\psi$. Also $(\nabla \times \mathbf{B}{)}_y=-{\nabla }^{2}A$, and henceThe buoyancy perturbation is $g{\alpha }_T(T-T_0)\hat{\mathbf{z}}$, whose curl in the $y$ direction is $-g{\alpha }_T(T-T_0{)}_x$. Taking the curl of the momentum equation and multiplying by minus one eliminates pressure and therefore givesWe have restored ${\mu }_0$ in this dimensional expression. The printed magnetic force uses units in which ${\mu }_0=1$, or equivalently a correspondingly rescaled field.

Use length scale $d$, time scale $d^2/\kappa$, velocity scale $\kappa /d$, stream-function scale $\kappa$, magnetic flux-function scale $B_{0}d$, and temperature perturbation scale $\Delta T$. Thus the dimensionless total flux function isHere $x,z$ now denote coordinates divided by $d$, and the same $T$ notation in the second identity denotes the physical temperature. The dimensionless parameters for thermal-diffusion scaling of planar magnetoconvection areThese are the Rayleigh number, Chandrasekhar number, Prandtl number and magnetic-to-thermal diffusivity ratio. The nondimensional magnetic-force coefficient is $B_0^2{d}^2/({\mu }_0{\rho }_r{\kappa }^2)=\sigma \zeta Q$, so omitting the factor $\zeta$ would use a different magnetic parameter.

The resistive induction equation becomes $A_t+J(\psi ,A)=\zeta {\nabla }^{2}A$. Since $J(\psi ,x)=-{\psi }_z$, its perturbation equation has source $+{\psi }_z$. Similarly, advection of the conductive temperature gradient gives $J(\psi ,-z)=-{\psi }_x$, producing the source $+{\psi }_x$ in the perturbation heat equation. The resulting nonlinear equations are thereforeThis derives both forcing signs from the background fields, rather than selecting them by a stream-function convention after the fact.

For the specified roll mode put $K=\pi /\lambda$ and $p_L=K^2+{\pi }^2$, the positive eigenvalue of minus the Laplacian on each spatial factor. Writeand take $\psi =(p_L/K)a{f}_s$, $\Theta =b{f}_T$, $\chi =\lambda c{f}_A$, with amplitudes depending on $\tau =p_{L}t$. These factors satisfy the printed boundary conditions: sine factors make the normal velocity and $B_x=-{\chi }_z$ vanish at the required faces; the second derivatives of the velocity factors give the stress-free conditions; $\Theta$ vanishes at the top and bottom and ${\Theta }_x$ vanishes at the sides.

Discard quadratic perturbation terms. Since ${\nabla }^{2}A={\nabla }^2\chi$, the magnetic force linearizes to $J(x,{\nabla }^2\chi )=({\nabla }^2\chi {)}_z$. The needed derivatives areThe heat and induction equations give $b^{\prime }=a-b$ and $c^{\prime }=a-\zeta c$. In the vorticity equation, its left side is $-(p_L^3/K)a^{\prime }{f}_s$ and its three right-side contributions are $\sigma [(p_L^3/K)a-RKb+\zeta Q{p}_L\pi \lambda c]f_s$. Consequently, withthe three-amplitude vertical-field magnetoconvection system isPrimes now mean $d/d\tau$. They are genuine time derivatives, as the original PDF shows.

For its linear stability, seek amplitudes proportional to $e^{s\tau }$. Eliminating $b,c$ or taking the determinant giveswhereAssume the physical range $\sigma >0$, $\zeta >0$, $q\ge 0$. The Routh-Hurwitz stability criterion gives strict decay exactly when $A_2>0$, $A_3>0$ and $A_1{A}_2>A_3$. At $r=0$ these hold. As $r$ increases, the first loss of stability can occur through a zero eigenvalue or a conjugate imaginary pair.

The stationary threshold isAt this value, $P(s)=s(s^2+A_{1}s+C)$ withIf $C>0$, the other two roots have negative real parts and the zero root is simple. Its derivative with respect to $r$ is $ds/dr=\sigma \zeta /C>0$. This is stationary onset; the reflection symmetry of the roll permits a generic pitchfork bifurcation. The linear calculation does not determine whether the nonlinear branch is supercritical or subcritical.

For oscillatory onset set $s=i\omega$, $\omega \ne 0$. Separating real and imaginary parts gives ${\omega }^2=A_2$ and $A_1{A}_2=A_3$. Solving the latter for $r$ yieldsA genuine Hopf bifurcation onset requires ${\omega }_H^2>0$, soAt this threshold $P(s)=(s+A_1)(s^2+{\omega }_H^2)$, giving the third root $-A_1<0$. The pair crosses transversely: differentiating $P(s,r)=0$ at $i{\omega }_H$ gives

The first instability of the conductive state is therefore stationary at $r_s$ if $\zeta \ge 1$, or if $0<\zeta <1$ and $q<q_{\ast }$. It is oscillatory at $r_H<r_s$ if $0<\zeta <1$ and $q>q_{\ast }$. At $q=q_{\ast }$, the two thresholds meet and $P(s)=s^2(s+A_1)$: the frequency is zero and neither a simple pitchfork nor a nonzero-frequency Hopf test applies. The oscillatory marginality in vertical-field magnetoconvection formula must not be used where its frequency squared is negative.

For conversion back to physical parameters, $R_{s,H}={\lambda }^2{p}_L^3{r}_{s,H}/{\pi }^2$, while the dimensional Hopf frequency is $(\kappa p_L/d^2){\omega }_H$. These are onset conditions for the specified spatial mode. Identifying the globally first mode would additionally require comparison over the allowed spatial modes. Nonlinear saturation and the stability of the bifurcating finite-amplitude states are not determined by this linear three-amplitude system alone.

The basic state has velocity fieldand buoyancy $B=N^{2}z$. For disturbances independent of $y$, the linearized equations areSubstituting a plane wave proportional to $\exp [i(kx+mz-\omega t)]$ and eliminating $p$, $v$, and $b$ gives the dispersion relationThus the requested coefficients are

An instability exists precisely when some wavenumber pair makes ${\omega }^2<0$. Since the Brunt–Väisälä frequency satisfies $N^2>0$, this is possible exactly whenThe basic relative vorticity is $\Lambda {\mathbf{e}}_z$, so its absolute vorticity is $(f+\Lambda ){\mathbf{e}}_z$. Its Ertel potential vorticity is thereforeThe instability criterion can consequently be written as $fQ<0$: the vertical absolute vorticity has the opposite sign to the planetary vorticity. This is inertial instability, approached most directly by disturbances with $|m/k|\gg 1$.

The replicator equation says that a strategy grows in frequency when its expected payoff exceeds the population's current mean payoff. It is an ordinary differential equation for a composition, rather than for total population size. The subtraction of the mean is what makes relative success matter: even a type growing in absolute number can shrink as a fraction if other types grow faster.

Let $x_i(t)$ be the frequency of pure strategy $i$, with $n$ available pure strategies andIn a symmetric bimatrix game, let the payoff matrix $A$ have entry $a_{ij}$ equal to the payoff to strategy $i$ against $j$. The other player's payoff in that encounter is $a_{ji}$: symmetry between player roles does not require $A=A^T$. Uniform random encounters give the expected payoffs and their population meanAfter choosing the time unit so that the growth-rate scale is one, the continuous-time replicator equation isHere a payoff is measured on a scale proportional to net growth per unit time, not merely as an arbitrary preference ranking. More generally one can specify locally Lipschitz frequency-dependent payoffs $f_i(x)$ directly; the same equation with $\bar{f}={\sum }_i{x}_i{f}_i$ then defines a unique frequency trajectory. The linear pairwise model and its reproductive interpretation are given in Taylor and Jonker (1978), sections 2–3; the formulas here also follow by the following direct normalization.

Suppose the number $N_i$ of type $i$ obeys ${\dot{N}}_i=N_i[b(t)+f_i(x)]$, where the baseline $b(t)$ is identical for all types. Set $N={\sum }_i{N}_i>0$ and $x_i=N_i/N$. The quotient rule givesThus the common baseline cancels. Adding the same constant to every entry of $A$ leaves the replicator equation unchanged; multiplying $A$ by a positive constant changes only the speed along trajectories. The same payoff matrix can therefore describe frequency evolution at different growth-rate baselines and time scales.

The probability simplex is invariant in forward time. On ${\sum }_i{x}_i=1$,Normalization alone would not exclude negative frequencies. Nonnegativity follows as well: along a solution,An absent type remains absent, and a positive frequency remains positive at every finite time. For the fixed payoff matrix model, bounded rates on the compact probability simplex give global forward solutions. Frequencies can still tend to zero as $t\to \infty$. Each face is invariant, so this model has no spontaneous introduction of new pure strategies; mutation or outside introductions would require additional terms. For two present types the logarithm of their frequency ratio satisfies $\frac{d}{dt}\log (x_i/x_j)=f_i-f_j$, directly expressing the accumulated payoff advantage.

For two pure strategies, write the composition as $(x,1-x)$, where $x$ is the frequency of strategy 1, and letTheir payoff difference and the resulting scalar ordinary differential equation areIn the interior $0<x<1$, the sign of the payoff difference determines the direction of motion. The endpoints are always equilibrium points, and an additional interior equilibrium point exists precisely when the affine payoff difference has a zero strictly between them. At a simple interior zero $x_{\ast }$, $F^{\prime }(x_{\ast })=x_{\ast }(1-x_{\ast })\Delta f^{\prime }(x_{\ast })$: a negative value gives asymptotic stability, a positive value gives instability. At the endpoints $F^{\prime }(0)=b-d$ and $F^{\prime }(1)=c-a$, with stability understood relative to $[0,1]$. A zero derivative requires examining the actual flow rather than this strict sign test.

For an original coordination game, imagine two conventions that reward matching, with convention 1 giving the larger matching reward. ChooseThe crossing is $x_{\ast }=2/5$. Below it, convention 1 loses frequency; above it, convention 1 gains frequency. The endpoint equilibrium points $0$ and $1$ have one-sided asymptotic stability, while $2/5$ is unstable: the corresponding derivatives are $-2,-3$ and $6/5$, respectively. Every initial $0<x(0)<2/5$ tends to $0$, every initial $2/5<x(0)<1$ tends to $1$, and exactly $x(0)=2/5$ stays there. Even the convention with the larger matching reward can disappear if it starts below the threshold. This is bistability, not selection of a universally best convention.

The compositions giving a symmetric Nash equilibrium in this coordination game are the two pure conventions and the mixed composition $(2/5,3/5)$. Both pure conventions are evolutionarily stable strategies because a mutant convention has strictly lower payoff against the pure resident. The mixed symmetric Nash equilibrium is not an evolutionarily stable strategy: against that resident both conventions tie, but after a pure convention-1 mutant is introduced its payoff advantage is $5(x-2/5)>0$. Equal payoffs at a symmetric Nash equilibrium need not provide resistance to invasion or asymptotic stability.

For an original coexistence model, choose payoffs making each convention advantageous when sufficiently rare:Here $x_{\ast }=3/4$ attracts every initial composition with both types present: $\dot{x}>0$ below it and $\dot{x}<0$ above it. The interior derivative is $F^{\prime }(3/4)=-3/4$, while $F^{\prime }(0)=3$ and $F^{\prime }(1)=1$ make both endpoints unstable to introduction of the other type. The endpoints nevertheless remain stationary if the other type is exactly absent; the attraction statement excludes those boundary initial states.

The sole symmetric Nash equilibrium in this coexistence model is $p=(3/4,1/4)$. It is also an evolutionarily stable strategy. For any different mixed strategy $q=(y,1-y)$, the first comparison in the two-condition criterion for evolutionary stability ties because $Ap=(3/2,3/2)$. Its second comparison is strictly positive:Thus restoring frequency feedback can protect a mixture even though all present types earn equal payoffs at the mixture. Coexistence here means different individuals using different pure strategies; it does not require every individual to randomize with a mixed strategy.

Frequency evolution need not increase the mean payoff. In the coexistence example $\bar{f}=5x-4x^2$, whose maximum occurs at $x=5/8$, whereas the attracting composition is $3/4$. For $5/8<x<3/4$, the frequency $x$ increases while $\frac{d}{dt}\bar{f}=(5-8x)\dot{x}<0$. Individual payoff advantages drive the equation; maximizing collective mean payoff is a different optimization problem.

More generally, a rest point only equalizes payoffs among present types: $x_i>0$ implies $(Ax{)}_i=x^{T}Ax$. A symmetric Nash equilibrium additionally requires $(Ax{)}_i\le x^{T}Ax$ for every absent type. All vertices are rest points even when an absent type could invade. For finite symmetric bimatrix games, a rest point with Lyapunov stability on the full probability simplex is a symmetric Nash equilibrium; an interior trajectory that converges has a symmetric Nash equilibrium as its limit. These are conditional results, not a convergence theorem for every trajectory. See Cressman and Tao (2014), Theorem 1.

The rock paper scissors payoff matrix gives an explicit obstruction to universal convergence:Since $A^T=-A$, $x^{T}Ax=0$, and in the interior $\frac{d}{dt}\log (x_1{x}_2{x}_3)={\sum }_i(Ax{)}_i=0$. Thus $x_1{x}_2{x}_3$ is conserved. Its maximum on the probability simplex occurs only at $(1/3,1/3,1/3)$, the unique symmetric Nash equilibrium, so an interior trajectory starting elsewhere cannot converge to that composition. Every other interior product level is a closed curve without a rest point, giving a periodic orbit. The existence of these cycles is also discussed in Cressman and Tao (2014), Example 1. The useful lesson is to inspect the specified payoff feedback and dynamical system, rather than infer convergence from the existence of an evolutionarily stable strategy or a Nash equilibrium.